An important industrial route to extremely pure acetic acid is the reaction of methanol with carbon monoxide: h5sil950 Use bond energies to calculate the heat of reaction.

Answer:

1. Which people are in the control group? The people who received the mint without the secret ingredient

(Group B) would be the control group.

2. What is the independent variable? Secret ingredient in the breath mint

3. What is the dependent variable? Amount of breath odor (or bad breath)

4. What should Mr. Krabs’ conclusion be? The breath mint with the secret ingredient appears to reduce the

amount of breath odor more than half the time, but it is not 100% effective.

5. Why do you think 10 people in group B reported fresher breath? This may be due to the placebo effect.

Answer:

23.28 g of O2.

Explanation:

We'll begin by calculating the mass of hexane. This can obtain as follow:

Volume of hexane = 10 mL

Density of hexane = 0.66 g/mL

Mass of hexane =?

Density = mass /volume

0.66 = mass of hexane /10

Cross multiply

Mass of hexane = 0.66 x 10

Mass of hexane = 6.6 g

Next, we shall write the balanced equation for the reaction. This is given below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation. This can be obtained as follow:

Molar mass of C6H14 = (12.01x6) + (1.008 x 14)

= 72.06 + 14.112

= 86.172 g/mol

Mass of C6H14 from the balanced equation = 2 x 86.172 = 172.344 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608 g

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Finally, we shall determine the mass of O2 needed to react with 10 mL (i.e 6.6 g) of hexane, C6H14. This can be obtained as follow:

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Therefore, 6.6 g of C6H14 will react with = (6.6 x 608)/172.344 = 23.28 g of O2.

Therefore, 23.28 g of O2 is needed for the reaction.

Answer:

A.

Explanation:

Neon 19 is shown on image A.

Diagram  A shows the correct model for an atom neon.

The arrangement of elements by their periodic properties and physical and chemical characteristic make a periodic table.

The chemical and physical properties of the group of the element belongs to Alkali Metals, Less dense, One loosely bound valence electron, Highly reactive, Low ionization energy, Low electronegativity

The lanthanides and actinides are also transition metals, these are softer and nonmetallic properties, all of these elements tend to have a shiny, metallic appearance, radioisotopes of other elements, actinides are radioactive.

Learn more about periodic table, here:

https://brainly.com/question/11155928

#SPJ2

Answer:

The concentration of monosodium phosphate is 0.1262M

Explanation:

The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2

To determine the pH you must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].

For H₂PO₄⁻ / HPO₄⁻ buffer:

pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]

As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:

7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]

0.2 = log [0.2] / [H₂PO₄⁻]

1.58489 = [0.2] / [H₂PO₄⁻]

[H₂PO₄⁻] = 0.1262M

Answer:

The correct answer is 1.06 M

Step-by-step explanation:

We have to calculate the molarity (M), which is:

M= moles solute/ 1 L solution.

The chemical formulae of manganese (II) nitrate is Mn(NO₃)₂. So, we first calculate its molecular weight (Mw) as follows:

Mw(Mn(NO₃)₂)= molar mass Mn + (2 x molar mass N) + (6 x molar mass N)= 55 g/mol + (2 x 14 g/mol) + (6 x 16 g/mol) = 179 g/mol

Then, with Mw we calculate the number of moles there is in the given mass of Mn(NO₃)₂:

moles Mn(NO₃)₂= mass/Mw= 23.8 g/(179 g/mol)= 0.133 mol

Now, we need the final volume in liters, so we convert the volume from mL to L:

125 mL x 1 L/1000 mL = 0.125 L

Finally, we divide the moles of Mn(NO₃)₂ into the volume in L, to obtain the molarity in mol/L:

M= 0.133 moles/0.125 L = 1.06 mol/L= 1.06 M

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

Answer:

The volume is 4.13793 L

Explanation:

Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:

[tex]density=\frac{mass}{volume}[/tex]

Density is a characteristic property of every body or substance.

The most commonly used units of density are [tex]\frac{kg}{m^{3} }[/tex] or [tex]\frac{g}{cm^{3} }[/tex] for solids, and [tex]\frac{kg}{L}[/tex] or [tex]\frac{g}{mL}[/tex] for liquids and gases.

In this case, you know:

Replacing:

[tex]0.87\frac{g}{mL} =\frac{3,600 g}{volume}[/tex]

Solving:

[tex]volume =\frac{3,600 g}{0.87\frac{g}{mL}}[/tex]

volume= 4,137.93 mL

Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L

The volume is 4.13793 L

Answer:

Option e.

Acid oxalate / oxalate

Explanation:

To determine the most efecttive buffer, we must look the pKas of the system.

If pKa value is similiar to buffer's pH, the variation of pH by  the addition of a small amount of strong acid or base is minimal. It would be a good buffer.

a. H₃PO₄ / H₂PO₄⁻   Ka 1 = 6.9×10⁻³

pKa = - log Ka → - log 6.9×10⁻³ = 2.16

b. H₂CO₃ / HCO₃⁻    Ka1 = 4.8×10⁻⁷

pKa = - log Ka → - log 4.8×10⁻⁷ = 6.32

c. HCO₃⁻ / CO₃⁻²    Ka2 = 4.8×10⁻¹¹

pKa = - log Ka → - log 4.8×10⁻¹¹ =  10.32

d.  H₂S / HS⁻    Ka1 = 8.9×10⁻⁸

pKa = - log Ka → - log 8.9×10⁻⁸ = 7.05

e. HC₂O₄⁻ / C₂O₄⁻²     Ka2 = 5.1×10⁻⁵

pKa = - log Ka → - log 5.1×10⁻⁵ = 4.29

pKa which is so close to the ph's value is e, acid oxalate / oxalate

Answer:

Solubility = [SO₄²⁻]

Explanation:

Solubility of Ag₂SO₄ can be understood seen its Ksp equilibrium:

Ag₂SO₄(s) ⇄ SO₄²⁻ + 2Ag⁺

Where 1 mole of silver sulfate dissolves producing 1 mole of sulfate ion and 2 moles of silver ion.

Solubility is defined as: "the amount of solid that can be dissolved per liter of solution".

In the dissolution process, the moles of Ag₂SO₄ dissolved are equal to moles of SO₄²⁻.

That means:

Answer:

a) Reactants are being consumed at the same rate they are being produced.

Explanation:

In a chemical reaction, chemical equilibrium is a state in which there is no change in concentration of the reactant and product. This means that as the reactants are consumed, they are also produced. Same thing with the products; immediately they are formed, they are consumed.

The option that accurately depicts a reaction at equilibrium is;

a) Reactants are being consumed at the same rate they are being produced.

Answer:

ΔH = ΔH₁ + ΔH₂ - ΔH₃

Explanation:

Given that:

1. A → 2B

2. B → C + D

3. E → 2D

Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.

To estimate the ΔH for the process A → 2C + E

We multiply 2 with equation 2 where (B → C + D)

2B → 2C + 2D ⇒ 2ΔH₂

Also, let's switch equation (3), such that we have,

2D → E -ΔH₃

The summation of all the equation result into :

A → 2C + E

where; ΔH = ΔH₁ + ΔH₂ - ΔH₃

Based on Hess law of constant heat summation, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3

According to Hess' law of constant heat summation, the heat of any reaction ΔH∘f for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction.

Thus, ΔH for the process A → 2C + E is calculated as follows:

The A → 2C + E can be obtained from the summation of the processes above:

A ---> 2B

2B ---> 2C + 2D

2D ---> E by reversing reaction step 3

Thus, ΔH = ΔH1 + ΔH2 - ΔH3

Therefore, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3

Learn more about enthalpy and Hess law at: https://brainly.com/question/11628413

Answer:

C. in the chemical wastebasket

Answer:

a. 1.728 moles.

b. 262.7g of Cr₂O₃ are required

Explanation:

Based on the reaction:

Cr₂O₃(s) + 3H₂S(g) → Cr₂S₃(s) + 3H₂O(l)

The important thing in the reaction is that 1 mole of Cr₂O₃ produce 1 mole of Cr₂S₃

a. To produce 346g of Cr₂S₃ we must know how many moles of Cr₂S₃ must be produced, and, as 1 mole of Cr₂O₃ produce 1 mole of Cr₂S₃ we can know moles of Cr₂O₃ that are required.

Moles of 346g Cr₂S₃ (Molar mass: 200.19g/mol):

346g Cr₂S₃ * (1mol / 200.19g) = 1.728 moles of Cr₂S₃

Based on the reaction, moles of Cr₂O₃ that are required are

b. Again, to conver the 1.728 moles of Cr₂O₃ to grams we must use molar mass of Cr₂O₃ (151.99g/mol):

1.728 moles Cr₂O₃ * (151.99g / mol) =

Answer:

4.02 gallons

Explanation:

From the illustration:

$0.750 USD = $1.00 CAD

Therefore,

$20 USD =  20 x 1/0.750 = $26.667 CAD

The cost of gas in the US is 2.21 USD per gallon.

20 USD will be able to get 20/2.21 = 9.05 gallons

The cost of gas across the border is 1.40 CAD per liter.

0.264 gallon = 1 liter

1 gallon = 1/0.264 = 3.788 liters

This means that 1 gallon = 3.788 x 1.40 = 5.303 CAD

Hence, 26.667 CAD will get 26.667/5.303 = 5.03 gallons

If he buys a $20 USD gas before crossing the border, Carl will get 9.05 gallons, but if he buys it after crossing the border, he will get 5.03 gallons. Therefore, if he buys the gas before crossing the border instead of buying it in Canada, he will get 9.05 - 5.03 = 4.02 gallons extra.

Answer:

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

Given :

0.00072 M solution of [tex]Ba(OH)_2[/tex] at [tex]25^oC[/tex] .

To Find :

The concentration of [tex]OH^-[/tex]and pOH .

Solution :

1 mole of [tex]Ba(OH)_2[/tex] gives 2 moles of [tex]OH^-[/tex] ions .

So , 0.00072 M mole of [tex]Ba(OH)_2[/tex] gives :

[tex][OH^-]=2 \times 0.00072\ M[/tex]

[tex][OH^-]=0.00144\ M[/tex]

[tex][OH^-]=1.44\times 10^{-3}\ M[/tex]

Now , pOH is given by :

[tex]pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84[/tex]

Hence , this is the required solution .

Answer:

1

Explanation:

Pyridine is a six membered hetrocyclic aromatic compound. Oxidation of the pyridine ring occurs at position-1.

Just like benzene, pyridine is stable towards mild oxidizing agents, however, pyridine is oxidized by peracetic acid to yield pyridine N-oxide.

Answer:

THE SPECIFIC HEAT OF THE METAL IS 28.69 J/g C

Explanation:

Mass of unknown metal = 30 g

Mass of water = 100 g

Initial Temperature of metal = 110 C

Initial temperature of water  = 200 C

Final temperature of the system = 25 C

Specific heat of water = 4200 J/ g C

Specific heat of metal = unknown

Heat lost = Heat gained\

M C Q = M C Q

30 * C * ( 25 - 110) = 100 * 4200 * (25 -200)

C = 100 * 4.18 * -175 / 30 * - 85

(the minus sign cancels out and we have):

C = 100 * 4.18 * 175 / 30 * 85

C = 73150 / 2550

C = 28.69 J/ g C

The specific heat of the unknown metal is hence 28.69 J/ g C

Answer:

[tex]30.0\frac{lb}{L}[/tex]

Explanation:

Hello,

In this case, since 454 g are equivalent to 1 pound and 1000 millilitres are equivalent to 1 liter, the required density is computed below by applying the corresponding conversion factor:

[tex]=13.6\frac{g}{mL} *\frac{1lb}{454g} *\frac{1000mL}{1L} \\\\=30.0\frac{lb}{L}[/tex]

Regards.

Answer:

3.25 kcal has the same amount of energy as 13600 J

Answer: Oxygen gas

Explanation: Gradpoint

Answer:

The neutral organic product is Imine.

Explanation:

From the image attached below, we will see an organic compound the question intended us to work out.

The proposed reaction is between cyclohexane and a stick structure of butylamine to yield imine as the product. The stepwise process and the mechanism carried out to get the product  of this reaction is also shown in the image attached below.

Answer:

no

Explanation:

light objects float

Answer:

See explanation

Explanation:

For sulphur dioxide;

3.53 g of sulphur combines with 3.52 g of oxygen

1.00 g of sulphur combines with 1.00 × 3.52/ 3.53 = 0.997 g

For sulphur trioxide

6.00 of sulphur combines with 9.00 g of oxygen

1.00 g of sulphur combines with 1.00 × 9.00/6.00 =1.5 g

Ratio of mass of oxygen;

1 : 1.5

The law of multiple proportions states that, if two elements A and B combine to form more than one chemical compounds, then the various masses of element A which combines separately with a fixed mass of element B are in simple multiple ratio.

We can see that the mass of oxygen which combines separately with sulphur in sulphur dioxide and sulphur trioxide are in simple ratio of 1:1.5.

Answer:

a. 8.31 x 10-5 s-1

Explanation:

The general first-order reaction is:

ln[A] = ln[A]₀ -kt

Where [A] is acutal concentration of reactant, initial concentration is [A]₀, k is rate constant and t is time pass

And the equation of the half-life, t 1/2, is:

[tex]t_{1/2} = \frac{ln 2}{K}[/tex]

Has half-life is 139min:

139min * (60s / 1min) = 8340s

[tex]t_{1/2} = \frac{ln 2}{8340s}[/tex]

Half-life is 8.31x10⁻⁵ s⁻¹

Answer:

O sólido tem densidade mais alta em comparação ao líquido.

Explicação:

A densidade da amostra sólida é maior do que a densidade do líquido porque há pouco espaço entre as partículas do sólido. A densidade tem relação inversa com o volume de uma substância, se uma substância ocupa mais espaço então sua densidade é menor, enquanto se a substância ocupa menos espaço então tem maior densidade. As substâncias sólidas ocupam menos espaço em comparação com as substâncias líquidas, então podemos dizer que a densidade do sólido é maior do que as substâncias líquidas.

Answer:

244/242 alright

Explanation:

yesss

Answer:

244/242 or 1.00826446 g/mL

Explanation:

Density can be calculated using the following formula.

d=m/v

where m is the mass and v is the volume.

The mass is 244 grams and the volume is 242 milliliters.

m= 244 g

v= 242 mL

Substitute the values into the formula.

d= 244 g / 242 mL

Divide 244 g by 242 mL.

d= 244/242 g/mL

d= 1.00826446 g/mL

The density is 244/242 or 1.00826446 grams per milliliter.

Answer:

A. Cars that are missing paint can rust over time shows a chemical property of rusting.

4Fe + 3O² + 6H²O → 4Fe(OH)³.

(Iron) (Air) (Water) (Rust)

Note: The powers in equation should be written below .

Hope it helps.